Question

розв'яжіть за теоремою Вієта:

1) 2х²+13х+6=0
2) 7х²-6х+2=0
3) 4х²+12х+9=0​

Answer 1

$2 {x}^{2} + 13x + 6 = 0$

$y_1y_2 = c \times a = 2 \times 6 = 12 \\ y_1 + y_2 = - 13 \\ y_1 = - 12 \\ y_2 = - 1 \\ x_1 = \frac{y_1}{a} = \frac{ - 12}{2} = - 6 \\ x_2 = \frac{y_2}{a} = \frac{ - 1}{2} = - 0.5$

$7 {x}^{2} - 6x + 2 = 0 \\ y_1y_2 = 14 \\ y_1 + y_2 = 6 \\ y ∉ R\\x ∉ R$

$4 {x}^{2} + 12x + 9 = 0 \\ y_1y_2 = 36 \\ y_1 + y_2 = - 12 \\ y_1 = - 9 \\ y_2 = - 4 \\ x_1 = - 4.5 \\ x_2 = - 2$